Minimum Substring Partition of Equal Character Frequency

Minimum Substring Partition of Equal Character Frequency

What is this problem about?

The Minimum Substring Partition of Equal Character Frequency problem asks you to partition a string into the minimum number of substrings such that each substring has equal frequency of all characters that appear in it. For example, "aab" has frequencies {a:2, b:1} — unequal — while "aabb" has {a:2, b:2} — equal. This Minimum Substring Partition of Equal Character Frequency coding problem blends dynamic programming with frequency counting.

Why is this asked in interviews?

Microsoft, Morgan Stanley, and Amazon ask this because it requires combining DP (for optimal partitioning) with frequency analysis. Checking whether a substring has balanced character frequencies is the inner condition, and the DP builds optimal partitions on top. The hash table, counting, string, and dynamic programming interview pattern are all relevant.

Algorithmic pattern used

DP with frequency checking. Define dp[i] = minimum number of partitions for the first i characters. For each ending position i, try all starting positions j < i. For the substring s[j..i-1], check if all character frequencies are equal (using a frequency counter). If valid, update dp[i] = min(dp[i], dp[j] + 1). Preprocessing character counts speeds up the inner check.

Example explanation

String: "fabccba".

• s[0..6] = "fabccba": freq {f:1, a:2, b:2, c:2} — not equal.
• s[0..0] = "f": freq {f:1} — equal (single char, vacuously). dp[1] = dp[0]+1 = 1.
• s[1..6] = "abccba": freq {a:2,b:2,c:2} — equal. dp[7] = dp[1]+1 = 2. Minimum partitions = 2.

Common mistakes candidates make

• Not recognizing that a single unique character always has "equal" frequency.
• Using O(n) frequency checks without memoization, causing TLE for long strings.
• Counting distinct characters incorrectly when checking if all frequencies are equal.
• Forgetting to initialize dp[0] = 0 (empty prefix needs zero partitions).

Interview preparation tip

DP problems with "valid partition" conditions follow a common template: iterate over partition end positions, check validity of each possible last segment, and update DP. The inner validation function — here "equal character frequency" — varies by problem but the DP framework is the same. Practice this template across different validity conditions (palindromes, balanced brackets, equal frequencies) to build versatile DP skills.